以8086汇编语言交换2个寄存器(16位) [英] swapping 2 registers in 8086 assembly language(16 bits)

问题描述

有人知道如何在不使用其他变量，寄存器，堆栈或任何其他存储位置的情况下交换2个寄存器的值吗?谢谢！

Does someone know how to swap the values of 2 registers without using another variable, register, stack, or any other storage location? thanks!

就像交换AX，BX.

推荐答案

您可以使用一些数学运算来完成.我可以给你个主意.希望对您有帮助！

You can do it using some mathematical operation. I can give you an idea. Hope it helps!

我遵循了以下C代码:

int i=10; j=20
i=i+j;
j=i-j;
i=i-j;

---

mov ax,10
mov bx,20
add ax,bx  
//mov command to copy data from accumulator to ax, I forgot the statement, now ax=30
sub bx,ax //accumulator vil b 10
//mov command to copy data from accumulator to bx, I forgot the statement now 
sub ax,bx //accumulator vil b 20
//mov command to copy data from accumulator to ax, I forgot the statement now 

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